Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which of the following are isostructural pairs ? A. ${\text{SO}}_4^{2-}$ and ${\text{CrO}}_4^{2-}$ B. ${\text{SiCl}}_4$ and ${\text{TiCl}}_4$ C. ${\text{NH}}_3$ and ${\text{NO}}_3^{-}$ D. ${\text{BCl}}_3$ and ${\text{BrCl}}_3$

When we talk about molecules being isostructural, we are strictly referring to their three-dimensional geometric shape. The prefix iso- means "same," and structural refers to the spatial arrangement of the atoms.

To determine the shape of any molecule, we rely on VSEPR (Valence Shell Electron Pair Repulsion) Theory. The golden rule here is to calculate the Steric Number (SN), which is the sum of the number of sigma bonds (bond pairs) and the number of lone pairs on the central atom.

While the steric number gives us the electron geometry, the actual molecular shape is determined only by the positions of the atoms. Lone pairs are invisible "ghosts" that push the atoms around but aren't seen in the final shape.

The Sulfate and Chromate Twins

Let's evaluate the first pair: ${\text{SO}}_4^{2-}$ and ${\text{CrO}}_4^{2-}$.

In the sulfate ion (${\text{SO}}_4^{2-}$), sulfur is the central atom. It forms four sigma bonds with four oxygen atoms and has zero lone pairs. A steric number of 4 with no lone pairs corresponds to an $s{p}^3$ hybridization and a perfect tetrahedral shape.

Now, look at the chromate ion (${\text{CrO}}_4^{2-}$). Chromium is a d-block transition metal, which might seem intimidating. However, in this $+6$ oxidation state, it uses its valence electrons to form four sigma bonds with oxygen, leaving no lone pairs. Just like sulfate, it adopts a tetrahedral geometry. Pair A is isostructural!

The Tetrachloride Siblings

Next up: ${\text{SiCl}}_4$ and ${\text{TiCl}}_4$.

Silicon is a group 14 element with four valence electrons. It forms four single bonds with chlorine atoms, leaving zero lone pairs. This gives us a steric number of 4, resulting in a tetrahedral shape.

Titanium, another d-block element, is in group 4. In ${\text{TiCl}}_4$, it is in a $+4$ oxidation state, meaning it has used all four of its valence electrons for bonding. Zero lone pairs and four bond pairs again yield a tetrahedral shape. Pair B is also isostructural!

The Nitrogen Misfits

Let's examine ${\text{NH}}_3$ and ${\text{NO}}_3^{-}$.

In ammonia (${\text{NH}}_3$), nitrogen has five valence electrons. It uses three to bond with hydrogen, leaving one lone pair. The steric number is $3+1=4$ ($s{p}^3$ hybridized). However, because one position is occupied by a lone pair, the shape is trigonal pyramidal.

In the nitrate ion (${\text{NO}}_3^{-}$), nitrogen forms three sigma bonds (one is a double bond, but it counts as one domain) and has zero lone pairs. The steric number is 3 ($s{p}^2$ hybridized), resulting in a flat trigonal planar shape. A pyramid and a plane are definitely not the same. Pair C is not isostructural.

The Halogen Oddballs

Finally, we look at ${\text{BCl}}_3$ and ${\text{BrCl}}_3$.

Boron in ${\text{BCl}}_3$ has three valence electrons and forms three bonds. With zero lone pairs, its steric number is 3, giving a trigonal planar shape.

Bromine in ${\text{BrCl}}_3$ is a halogen with seven valence electrons. It uses three to bond with chlorine, leaving four electrons, which pair up to form two lone pairs. The steric number is $3+2=5$ ($s{p}^{3}d$ hybridized). In a trigonal bipyramidal electron geometry, the two lone pairs occupy the equatorial positions to minimize repulsion, forcing the three chlorine atoms into a T-shaped geometry. A plane and a T-shape do not match. Pair D is not isostructural.

After a thorough VSEPR analysis, we found that only Pair A and Pair B consist of molecules with identical shapes (tetrahedral). Therefore, the correct answer is (a) A and B only.